Question

Fie ecuatia x²-2mx-1=0. Sa se determine m€R, astfel incat ecuatia sa indeplineasca conditia:
√|x1| + √|x2| = 2

Answer 1

x^2-2mx-1=0
delta=(2m)^2-4*1*(-1)=4m^2+4=4(m^2+1).
x1,2=(2m+-rad(4(m^2+1)))/2=(2m+-2rad(m^2+1))/2=m+-rad(m^2+1).

Pe baza relatiilor lui Viete avem x1+x2=-b/a=2m si x1*x2=c/a=-1.

4=(rad(|x1|)+rad(|x2|))^2=|x1|+|x2|+2rad(|x1*x2|)=|x1|+|x2|+2. => |x1|+|x2|=2.

Daca x1>=x2: |x1|+|x2|=2 <=> m+rad(m^2+1)-(m-rad(m^2+1))=2 <=> 2rad(m^2+1)=2 <=> rad(m^2+1)=1 <=> m^2+1=1 <=> m^2=0 <=> m=0.

Daca x1<x2: -(m-rad(m^2+1))+m+rad(m^2+1)=2 <=> 2rad(m^2+1)=2 <=> rad(m^2+1)=1 <=> m^2+1=1 <=> m^2=0 <=> m=0.

In concluzie, m=0.